How do you use substitution to solve the pair of equations: y=2x-3y=2x−3 and x^2+y^2=16x2+y2=16?

1 Answer
Alan P.
Dec 6, 2016

(x,y)~~color(green)(""(2.885,2770)) or color(green)(""(-0.485,-3.970))(x,y)≈(2.885,2770)or(−0.485,−3.970)

Explanation:

Given
[1]color(white)("XXX")y=2x-3XXXy=2x−3
[2]color(white)("XXX")x^2+y^2=16XXXx2+y2=16
Using [1] we know that we can substitute (2x-3)(2x−3) for yy in [2]
[3]color(white)("XXX")x^2+(2x-3)^2=16XXXx2+(2x−3)2=16

[4]color(white)("XXX")rarr x^2+4x^2-12x+9=16XXX→x2+4x2−12x+9=16

[5]color(white)("XXX")rarr5x^2-12x-7=0XXX→5x2−12x−7=0

Using the quadratic formula:
[6]color(white)("XXX")x=(12+-sqrt((-12)^2-4 * 5 * (-7)))/(2 * 5)XXXx=12±√(−12)2−4⋅5⋅(−7)2⋅5

simplifying (and using a calculator)
color(white)("XXX"){:(x~~2.885," or ",x~~-0.485):}
using [1] rarr
color(white)("XXX"){:(y~~2.770," or ", y~~-3.970):}

The image of the given relations (below) indicates that these results are reasonable.
enter image source here

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